Degenerate Hilbert-Schmidt operators

Question

Let us define a Hilbert Schmidt operator $A:L_2[a,b]\to L_2[a,b]$ by $$A\varphi:=\int_{[a,b]} K(s,t)\varphi(t)d\mu_t$$where $\mu_t$ is the linear Lebesgue measure. A degenerate case is represented by a $K$ having the form $\sum_{i=1}^n P_i(s)Q_i(t)$ where $\forall i=1,\ldots,n\quad P_i,Q_i\in L_2[a,b]$ and such $P_i, Q_i$ can be assumed to be linearly independent, without loss of generality. I read (p. 469 of Kolmogorov-Fomin's here) that any operator $A$ is the limit of a succession of degenerate cases. How can it be proved? I thank you all very much!

Answer 1

Let $(\phi_i)_i$ be an orthonormal basis of $L^2(a,b)$. Define the basis $(\Phi_{ij})_{i,j}$ of $L^2([a,b] \times [a,b])$ via tensor products, $\Phi_{ij}(s,t) = \phi_i(s) \phi_j(t)$. Now expand the kernel $K$ in this basis, $$ K = \sum_{i,j} \alpha_{ij} \Phi_{ij}. $$ Truncate the expansion and you have a degenerate approximation: $$ K_N(s,t) = \sum_{i,j \le N} \alpha_{ij} \Phi_{ij}(s,t) = \sum_{i,j \le N} \alpha_{ij} \phi_{i}(s)\phi_j(t) \, . $$